Menu Close

Question-25702




Question Number 25702 by ajfour last updated on 13/Dec/17
Commented by ajfour last updated on 13/Dec/17
solution to Q. 25700
$${solution}\:{to}\:{Q}.\:\mathrm{25700} \\ $$
Answered by ajfour last updated on 13/Dec/17
2x=1cm  x+y+z =2cm       ...(i)  cot θ=((y+z)/R)  , and ..(ii)  sin θ =((R−r)/(R+r)) =(x/(x+y+z))       ...(iii)  ⇒sin θ= ((R−r)/(R+r))=(1/4)  or   5r=3R     ⇒ cot θ =(√(15))   considering (ii):     (√(15)) =(3/(2R))  ⇒ R= (3/(2(√(15)))) =((√(15))/(10)) .      r=(3/5)×((√(15))/(10)) =((3(√(15)))/(50)) .
$$\mathrm{2}{x}=\mathrm{1}{cm} \\ $$$${x}+{y}+{z}\:=\mathrm{2}{cm}\:\:\:\:\:\:\:…\left({i}\right) \\ $$$$\mathrm{cot}\:\theta=\frac{{y}+{z}}{{R}}\:\:,\:{and}\:..\left({ii}\right) \\ $$$$\mathrm{sin}\:\theta\:=\frac{{R}−{r}}{{R}+{r}}\:=\frac{{x}}{{x}+{y}+{z}}\:\:\:\:\:\:\:…\left({iii}\right) \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\:\frac{{R}−{r}}{{R}+{r}}=\frac{\mathrm{1}}{\mathrm{4}}\:\:{or}\:\:\:\mathrm{5}{r}=\mathrm{3}{R}\:\:\: \\ $$$$\Rightarrow\:\mathrm{cot}\:\theta\:=\sqrt{\mathrm{15}}\: \\ $$$${considering}\:\left({ii}\right): \\ $$$$\:\:\:\sqrt{\mathrm{15}}\:=\frac{\mathrm{3}}{\mathrm{2}{R}}\:\:\Rightarrow\:{R}=\:\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{15}}}\:=\frac{\sqrt{\mathrm{15}}}{\mathrm{10}}\:. \\ $$$$\:\:\:\:{r}=\frac{\mathrm{3}}{\mathrm{5}}×\frac{\sqrt{\mathrm{15}}}{\mathrm{10}}\:=\frac{\mathrm{3}\sqrt{\mathrm{15}}}{\mathrm{50}}\:. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *